Function 0

\(f_0(x; a_0) = a_0\) where \(x \geq 0\)

In this model, \(a_0\) seems to vertically translate the function by specifying the y-intercept

Function 1

\(f_1(x; a_0,a_1) = a_0 + a_1x\) where \(x \geq 0\)

In this model, \(a_0\) seems to vertically translate the function by specifying the y-intercept

\(a_1\) seems to change the slope at the y-intercept.

Function 2

\(f_2(x; a_0,a_1,a_2) = a_0 + a_1x + a_2x^2\) where \(x \geq 0\)

In this model, \(a_0\) seems to vertically translate the function by specifying the y-intercept

\(a_1\) seems to change the slope at the y-intercept.

\(a_2\) seems to horizontally stretch the function. If negative, \(a_2\) seems to reflect the function over the x-axis

Function 3

\(f_3(x; a_0,a_1,a_2 ) = a_0 + a_1e^{-a_2x}\) where \(x \geq 0\)

In this model, \(a_0\) seems to vertically translate the function by specifying the y-intercept

\(a_1\) seems to change the tail of the exponential function. If \(a_1\) is negative, as x goes to infinity, y will be negative. If positive, as x goes to infinity, y will be positive.

\(a_2\) seems to horizontally stretch the function. If negative, \(a_2\) seems to reflect the function over the y-axis

Function 4

\(f_4(x; a_0,a_1,a_2) = a_0 + a_1x + a_2\ln(0.005x+1)\) where \(x \geq 0\)

In this model, \(a\) seems to vertically translate the function by specifying the y-intercept

\(b\) seems to change the slope at the y-intercept. If \(b\) is negative, as x goes to infinity, y will be negative. If positive, as x goes to infinity, y will be positive.

\(c\) will vertically stretch the function. If negative, the function will vertically reflect over the x-axis.

Function 5

\(f_5(x; a_0,a_1,a_2) = (a_0+a_1x)e^{-a_2x}\) where \(x \geq 0\)

In this model, \(a_0\) seems to vertically manipulate the function by specifying the y-intercept. While some of the function will translate, the end behavior of the function will remain constant

\(a_1\) seems to vertically stretch the function. If negative, \(a_1\) will vertically reflect over where \(a_0\) is at.

\(a_2\) seems to horizontally change the end behaviors of the function. If \(a_1\) and \(a_2\) is positive, each end will display in quadrant 1 and 3. If both negative, the ends will display in quadrant 2 and 4.

Function Zero

\(f_0(t; a_0) = a_0\) where \(t \geq 0\)

Given the function above, in order to make the model visually accurate, the parameters specified in the function must be values that allow the above to be so. For example, since \(f_0(x)\) must be equal to 100 when \(t = 0\), \(a_0\) is set to “100”.

# f_0(t; a_0)  =  a_0
f_0 <- function(t, a0 = 100){
  a0 + t*0  
}

par(mfrow = c(1,2),mar = c(2,2,3,0.25),oma = rep(0.5,4))
plot(t,y1,xlab = "Hour ", ylab = "Intensity(%) ", pch = 16,main = '')
lines(x,f_0(x),col = 2)
plot(t,y1,xlab = "Hour ", ylab = "Intensity(%) ", pch = 16, xlim  =  c(-10,80000),ylim  =  c(-10,120))
lines(x,f_0(x),col = 2)

Function One

\(f_1(t; a_0,a_1) = a_0 + a_1t\) where \(t \geq 0\)

Given the function above, in order to make the model visually accurate, the parameters specified in the function must be values that allow the above to be so. For example, since \(f_1(x)\) must be equal to 100 when \(t = 0\), \(a_0\) is set to “100”.

\(a_1\) is set to the positive value, “0.00015” since the curve of the LED data seems to be positive

# f_1(t; a_0,a_1)  =  a_0 + a_1t
f_1 <- function(t, a0  =  100, a1  =  0.00015){
  a0 + a1*t
}

par(mfrow = c(1,2),mar = c(2,2,3,0.25),oma = rep(0.5,4))
plot(t,y1,xlab = "Hour ", ylab = "Intensity(%) ", pch = 16,main = '')
lines(x,f_1(x),col = 2)
plot(t,y1,xlab = "Hour ", ylab = "Intensity(%) ", pch = 16, xlim  =  c(-10,80000),ylim  =  c(-10,120))
lines(x,f_1(x),col = 2)

Function Two

\(f_2(t; a_0,a_1) = a_0 + a_1t\) where \(t \geq 0\)

Given the function above, in order to make the model visually accurate, the parameters specified in the function must be values that allow the above to be so. For example, since \(f_2(x)\) must be equal to 100 when \(t = 0\), \(a_0\) is set to “100”.

\(a_1\) and \(a_2\) are set as “0.00035” and “-0.00000005” respectively such that the curve visually matches the data. Since \(a_2\) changes whether the parabola opens up or down, this is set to negative such that as the value of x increases, the value of y will first increase, then decrease.

# f_2(t; a_0,a_1)  =  a_0 + a_1t
f_2 <- function(t, a0  =  100, a1  =  0.00035, a2  =  -0.00000005){
  a0 + a1*t + a2*t^2
}

par(mfrow = c(1,2),mar = c(2,2,3,0.25),oma = rep(0.5,4))
plot(t,y1,xlab = "Hour ", ylab = "Intensity(%) ", pch = 16,main = '')
lines(x,f_2(x),col = 2)
plot(t,y1,xlab = "Hour ", ylab = "Intensity(%) ", pch = 16, xlim  =  c(-10,80000),ylim  =  c(-10,120))
lines(x,f_2(x),col = 2)

Function Three

\(f_3(t; a_0,a_1,a_2 ) = a_0 + a_1e^{-a_2t}\) where \(t \geq 0\)

Given the function above, in order to make the model visually accurate, the parameters specified in the function must be values that allow the above to be so. Since \(a_1\) and \(a_2\) are “-0.77” and “0.0005” respectively, \(a_0\) is set to “100.77” so that \(f_3(x)\) is equal to zero. \(a_1\) and \(a_2\) are set as such that the curve visually matches the data. \(a_1\) is set to a negative value, “-0.77” to reflect the fact that as the value of x decreases, \(f_3(x)\) decreases as well. Finally, \(a_2\) is set to “0.0005” to refine the intensity of the curve.

# f_3(t; a_0,a_1,a_2 )  =  a_0 + a_1e^{-a_2t}
f_3 <- function(t, a0  =  100.77, a1  =  -0.77, a2  =  0.0005){
  a0 + a1*exp(-a2*t)
}

par(mfrow = c(1,2),mar = c(2,2,3,0.25),oma = rep(0.5,4))
plot(t,y1,xlab = "Hour ", ylab = "Intensity(%) ", pch = 16,main = '')
lines(x,f_3(x),col = 2)
plot(t,y1,xlab = "Hour ", ylab = "Intensity(%) ", pch = 16, xlim  =  c(-10,80000),ylim  =  c(-10,120))
lines(x,f_3(x),col = 2)

Function Four

\(f_4(t; a_0,a_1,a_2) = a_0 + a_1t + a_2\ln(0.005t+1)\) where \(t \geq 0\)

Given the function above, in order to make the model visually accurate, the parameters specified in the function must be values that allow the above to be so. For example, since \(f_4(x)\) must be equal to 100 when \(t = 0\), \(a_0\) is set to “100”.

\(a_1\) and \(a_2\) are set as “0.000005” and “0.14” respectively such that the curve visually matches the data. \(a_1\) is set to a positive value so that as the x value increases, \(f_4(x)\) increases. \(a_2\) is set to a positive value so that as the the x value decreases, \(f_4(x)\) decreases.

# f_4(t; a_0,a_1,a_2)  =  a_0 + a_1t + a_2\ln(0.005t+1)
f_4 <- function(t, a0  =  100, a1  =  0.000005, a2  =  0.14){
  a0 + a1*t + a2*log(0.005*t + 1)
}

par(mfrow = c(1,2),mar = c(2,2,3,0.25),oma = rep(0.5,4))
plot(t,y1,xlab = "Hour ", ylab = "Intensity(%) ", pch = 16,main = '')
lines(x,f_4(x),col = 2)
plot(t,y1,xlab = "Hour ", ylab = "Intensity(%) ", pch = 16, xlim  =  c(-10,80000),ylim  =  c(-10,120))
lines(x,f_4(x),col = 2)

Function Five

\(f_5(t; a_0,a_1,a_2) = (a_0+a_1t)e^{-a_2t}\) where \(t \geq 0\)

Given the function above, in order to make the model visually accurate, the parameters specified in the function must be values that allow the above to be so. For example, since \(f_5(x)\) must be equal to 100 when \(t = 0\), \(a_0\) is set to “100”.

\(a_1\) and \(a_2\) are set as “-0.00374” and “-0.000042” respectively such that the curve visually matches the data. \(a_1\) and \(a_2\) play off each other such that as these values approach zero together, their curve becomes less intense and wider.

# f_5(t; a_0,a_1,a_2) = (a_0+a_1t)e^{-a_2t}
f_5 <- function(t, a0 = 100, a1 = -0.00374, a2 = -0.000042){
  (a0 + a1*t)*exp(-a2*t)
}

par(mfrow = c(1,2),mar = c(2,2,3,0.25),oma = rep(0.5,4))
plot(t,y1,xlab = "Hour ", ylab = "Intensity(%) ", pch = 16,main = '')
lines(x,f_5(x),col = 2)
plot(t,y1,xlab = "Hour ", ylab = "Intensity(%) ", pch = 16, xlim = c(-10,80000),ylim = c(-10,120))
lines(x,f_5(x),col = 2)

Manual calculation

Computational calculation

Uniroot approximates solving \(f_2\) given the interval. There is an additional negative solution that is not included.

f_2 <- function(t, a0  =  100, a1  =  0.00035, a2  =  -0.00000005){
  a0 + a1*t + a2*t^2
}

f_2a <- function(x){
  f_2(x) - 80
}

uniroot(f_2a, c(23000,24000))$root

## [1] 23803.94

f2at80 <- uniroot(f_2a, c(23000,24000))$root

Manual calculation

Computational calculation

Uniroot approximates solving \(f_5\) given the interval. There is an additional negative solution that is not included.

f_5 <- function(t, a0 = 100, a1 = -0.00374, a2 = -0.000042){
  (a0 + a1*t)*exp(-a2*t)
}

f_5a <- function(x){
  f_5(x) - 80
}

uniroot(f_5a, c(15000,20000))$root

## [1] 15654.55

f5at80 <- uniroot(f_5a, c(15000,20000))$root

From my completely subjective estimate, I will average model two and five to get an estimate on when the LED’s will decrease to 80 percent of original intensity.

mean(c(f2at80, f5at80))

## [1] 19729.25

The LED light bulb will “burn out” after 19,729.25 hours.